Example 6-11 A Ski Jump, Revisited
As in Example 6-10 (Section 6-6), a skier of mass m starts
at rest at the top of a ski jump ramp (Figure 6-31). (a) Use
an energy conservation equation to find an expression
for the skier’s speed as she flies off the ramp. There is
negligible friction between the skis and the ramp, and you
can ignore air resistance. (b) After the skier reaches the
ground at point P in Figure 6-32, she begins braking and
comes to a halt at point Q. Find an expression for the
magnitude of the constant friction force that acts on her
between points P and Q.
Skier, mass m
vi = 0
Speed vf = ?
H
D
P
h
L
Figure 6-31 ​Flying off the ramp and coming to a halt What is
the skier's speed as she leaves the ramp? How much friction is
required to bring her to a halt?
Set Up
While the skier is on the ramp, both a normal force
and the gravitational force act on her. The normal
force is always perpendicular to her motion, so it
does no work on her. So for part (a) of this problem,
which takes us from the beginning to the end of the
ramp, only the conservative force of gravity does
work and so we can use Equation 6-23. (There are
no springs, so there is no spring potential energy.)
Our goal for this part is to find the skier’s speed
vf as she leaves the ramp. In this part we take the
initial point i to be at the top of the ramp and the
final point f at the end of the ramp.
For part (b) we consider the skier’s entire motion
from the top of the ramp (which we again call point i)
to where she finally comes to rest at point Q. A
normal force also acts on her when she reaches the
ground, but again this force does no work on her.
Between points P and Q, a nonconservative friction
force also acts and does negative work on her (the
force is opposite to her motion). So for the motion as
a whole we must use Equation 6-26. Our goal for
part (b) is to find the magnitude fk of the friction force.
Conservation of mechanical energy:
Ki + Ui = Kf + Uf
n
(6-23)
Gravitational potential energy:
(6-14)
Ugrav = mgy
Change in mechanical energy when
nonconservative forces act::
Ki + Ui + Wnonconservative = Kf + Uf
n
mg
mg
on ramp
in midair
f
k
(6-26)
Kinetic energy:
1
K = mv 2
2
(6-8)
mg
mg
mg
on ramp
in midair
between
P and Q
fk
n
n
At the top of the ramp:
1
K i = mv 2i = 0
2
Ugrav, i = mgyi = mgH
mg
mg
mg
on ramp
in midair
between
P and Q
i
H
K Ugrav
At the end of the ramp:
1
K f = mv 2f
2
Ugrav, f = mgyf = mgD
Now substitute the energies into Equation 6-23 and
solve for vf.
Q
Friction force fk = ?
Equation 6-23:
Ki + Ui = Kf + Uf , so
1
0 + mgH = mv 2f + mgD
2
1
2
mv f = mgH - mgD = mg 1H - D2
2
v 2f = 2g1H - D2
vf = 22g1H - D2
K Ugrav
f
D
(b) Now the final position is at point Q, a distance
h below the low point of the ramp so yQ = 2h.
At this point the skier is again at rest, so vQ = 0.
The friction force fk acts opposite to her motion as
she moves a distance L, so this force does work 2fkL
on her.
At the top of the ramp:
1
K i = mv 2i = 0
2
Ugrav, i = mgyi = mgH
i
K
Ugrav
At point Q:
1
K Q = mv 2Q = 0
2
Ugrav,Q = mgyQ
= mg1 -h2 = -mgh
Ugrav
K
Q
Work done by the friction force, which points opposite
to the skier’s displacement:
Wnonconservative = fkL cos 180° = fkL(21)
= 2fkL
Now use Equation 6-26 to solve for fk.
Reflect
Equation 6-26:
Ki + Ui + Wnonconservative = KQ + UQ so
0 + mgH + (2fkL) = 0 + (2mgh)
fkL = mgH + mgh = mg (H + h)
H + h
b
fk = mga
L
Our answer for part (a) is the same one that we found in Example 6-10, where we used the expression Wgrav = 2DUgrav for
the work done by gravity. That’s as it should be, since we used that expression in Section 6-7 to help derive Equation 6-23.
Our result for fk in part (b) has the correct dimensions of force, the same as mg, since the dimensions of H + h (the
total vertical distance that the skier descends) cancel those of L (the skier’s stopping distance). Note that the smaller the
value of L compared to H + h, the greater the friction force required to bring the skier to a halt.
An alternative way to solve part (b) would be to treat the second part of the skier’s motion separately, with the
initial point at the end of the ramp [at height D, where the skier’s speed is 12g1H - D2 as found in part (a)] and the
final point is at point Q (at height 2h, where the skier’s speed is zero). You should try solving the problem this way using
Equation 6-26 and Wnonconservative = 2fkL; you should get the same answer for fk. Do you?